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Rashid [163]
3 years ago
14

Determine the volume, in liters, occupied by 0.015 molecules of oxygen at STp?

Chemistry
1 answer:
Arlecino [84]3 years ago
5 0

5.58 X 10^{-25} Litres is the volume, in liters, occupied by 0.015 molecules of oxygen at STP.

Explanation:

Data given:

molecules of oxygen = 0.015

number of moles of oxygen =?

temperature at STP = 273  K

Pressure at STP = 1 atm

volume = ?

R (gas constant) = 0.08201 L atm/mole K

to convert molecules to moles,

number of moles = \frac{molecules}{Avagadro's number}

number of moles = 2.49 x 10^{-26}

Applying the ideal gas law since the oxygen is at STP,

PV = nRT

rearranging the equation:

V = \frac{nRT}{P}

putting the values in the rearranged equation:

V = \frac{2.49 X10^{-26}   X 0.08201 X 273}{1}

V = 5.58 X 10^{-25} Litres.

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Estimate the freezing and normal boiling points of 0.25 m aqueous solutions of nh4no3 nicl3 al2so43
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The items are answered below and are numbered separately for each compound. 

The freezing point of impure solution is calculated through the equation,
     Tf = Tfw - (Kf)(m)

where Tf is the freezing point, Tfw is the freezing point of water, Kf is the freezing point constant and m is the molality. For water, Kf is equal to 1.86°C/m. In this regard, it is assumed that m as the unit of 0.25 is molarity.

1. NH4NO3
    Tf = 0°C - (1.86°C/m)(0.25 M)(2) = -0.93°C

2. NiCl3 
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3. Al2(SO4)3
      Tf = 0°C - (1.86 °C/m)(0.25 M)(5) = -2.325°C

For boiling points, 
    Tb = Tbw + (Kb)(m)
For water, Kb is equal to 0.51°C/m.

1. NH4NO3
     Tb = 100°C + (0.51°C/m)(0.25 M)(2) = 100.255°C

2. NiCl3
     Tb = 100°C + (0.51°C/m)(0.25 M)(4) = 100.51°C

3. Al2(SO4)3
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